Consider a curve y=y(x) in the first quadrant as shown in the figure. Let the area A₁ is twice the area A₂. Then the normal to the curve perpendicular to the line 2x-12y=15 does NOT pass through the point.

Solution

Maths sir · 4 steps

Differentiate the area condition with respect to x, and the curve turns out to be a parabola.

1. Write both areas. From the figure, A₁ = ∫₄ˣ y dx is the area under the curve, and A₂ = ∫₂ʸ x dy is the area between the curve and the y-axis. The condition A₁ = 2A₂ holds for every end point (x, y), so differentiate both sides with respect to x.

   y = 2xdy/dx ⇒ (2 dy)/y = dx/x ⇒ y² = kx

2. Fix the constant. The curve passes through (4, 2), so 4 = 4k and k = 1. The curve is y² = x.
3. Normal with slope −6. The line 2x - 12y = 15 has slope 1/6, so the normal must have slope -6. For y² = x the tangent slope is 1/2y, so the normal slope is -2y = -6, giving y = 3, x = 9. The normal is y - 3 = -6(x - 9), that is, y = 57 - 6x.

   Point	57 - 6x	On the normal?
   (6, 21)	57 - 36 = 21	yes
   (8, 9)	57 - 48 = 9	yes
   (10, −4)	57 - 60 = -3	no, since -3 ≠ -4
   (12, −15)	57 - 72 = -15	yes

4. So. The normal y = 57 - 6x passes through every given point except (10, -4), so the answer is (c).

The idea to keep. When an area condition holds for every end point, differentiate it with respect to that end point. The integral sign disappears and you get a differential equation for the curve.

Answer from the NTA final answer key. Solution by PrepWiser.